Úlohy - limita - limita v nevlastním bodě
Úloha
\(\lim_{x \to +\infty} x - \sqrt{x^2 - 1} = \)
- \(= \lim_{x \to +\infty} \frac {(x - \sqrt{x^2 - 1})(x + \sqrt{x^2 - 1})}{(x + \sqrt{x^2 - 1})} = \)
- \(= \lim_{x \to +\infty} \frac {x^2 - (x^2 - 1)}{(x + \sqrt{x^2 - 1})} = \)
- \(= \lim_{x \to +\infty} \frac {1}{(x + \sqrt{x^2 - 1})} = \)
- \(= 0 \)
Úloha
\(\lim_{x \to +\infty} (4x^3 - x^2 + x + 2) = \)
- \(= \lim_{x \to +\infty} 4x^3 \left( 1 - \frac {x^2}{4x^3} + \frac {x}{4x^3} + \frac {2}{4x^3} \right ) = \)
- \(= \lim_{x \to +\infty} 4x^3 \left( 1 - \frac {1}{4x} + \frac {1}{4x^2} + \frac {2}{4x^3} \right ) = \)
- \(= +\infty \)
Úloha
\(\lim_{x \to +\infty} (-4x^3 - x^2 + x + 2) = \)
- \(= \lim_{x \to +\infty} (-4x^3) \left( 1 + \frac {1}{4x} - \frac {1}{4x^2} - \frac {2}{4x^3} \right ) = \)
- \(= -\infty \)
Úloha
\(\lim_{x \to +\infty} \frac{2x^3 - x^2 + 5}{x^2 + x - 2} = \)
- \(= \lim_{x \to +\infty} \frac {x^3}{x^2} \cdot \frac{2 - \frac{1}{x} + \frac{5}{x^3}}{1 + \frac{1}{x} - \frac{2}{x^2}} = \)
- \(= \lim_{x \to +\infty} \frac {x^3}{x^2} \cdot \lim_{x \to +\infty} \frac{2 - \frac{1}{x} + \frac{5}{x^3}}{1 + \frac{1}{x} - \frac{2}{x^2}} = \)
- \(= +\infty \)
Úloha
\(\lim_{x \to -\infty} \frac{4x^3 - x + 2}{3x^3 + x^2 + x - 2} = \)
- \(= \lim_{x \to -\infty} \frac {x^3}{x^3} \cdot \frac{4 - \frac{1}{x^2} + \frac{2}{x^3}}{3 + \frac{1}{x} + \frac{1}{x^2} - \frac{2}{x^3}} = \)
- \(= \lim_{x \to -\infty} 1 \cdot \lim_{x \to -\infty} \frac{4 - \frac{1}{x^2} + \frac{2}{x^3}}{3 + \frac{1}{x} + \frac{1}{x^2} - \frac{2}{x^3}} = \)
- \(= \frac{4}{3} \)
Úloha
\(\lim_{x \to -\infty} \frac{x^2 - 2x + 5}{2x^3 - x^2 + 4} = \)
- \(= \lim_{x \to -\infty} \frac {x^2}{x^3} \cdot \frac{1 - \frac{2}{x} + \frac{5}{x^2}}{2 - \frac{1}{x} + \frac{4}{x^3}} = \)
- \(= \lim_{x \to -\infty} \frac {1}{x} \cdot \lim_{x \to -\infty} \frac{1 - \frac{2}{x} + \frac{5}{x^2}}{2 - \frac{1}{x} + \frac{4}{x^3}} = \)
- \(= 0 \)
Úloha
\(\lim_{x \to +\infty} \frac{\sqrt{x^2 - 1} + \sqrt{x^2 + 1}}{x} = \)
- \(= \lim_{x \to +\infty} \sqrt{\frac{x^2 - 1}{x^2}} + \sqrt{\frac{x^2 + 1}{x^2}} = \)
- \(= \lim_{x \to +\infty} \sqrt{1 - \frac{1}{x^2}} + \sqrt{1 + \frac{1}{x^2}} = \)
- \(= 2 \)
Úloha
\(\lim_{x \to +\infty} (\sqrt{x - 2} - \sqrt{x}) = \)
- \(= \lim_{x \to +\infty} (\sqrt{x - 2} - \sqrt{x})\cdot \frac {\sqrt{x - 2} + \sqrt{x}}{\sqrt{x - 2} + \sqrt{x}} = \)
- \(= \lim_{x \to +\infty} \frac{x - 2 - x}{\sqrt{x - 2} + \sqrt{x}} = \)
- \(= \lim_{x \to +\infty} \frac{-2}{\sqrt{x - 2} + \sqrt{x}} = \)
- \(= 0 \)
Úloha
\(\lim_{x \to +\infty} \frac{\sqrt{x^2 + 1} - x}{x + 1} = \)
- \(= \lim_{x \to +\infty} \frac {\sqrt{x^2 + 1} - x}{x + 1} \cdot \frac {\frac {1}{x}}{\frac{1}{x}} = \)
- \(= \lim_{x \to +\infty} \frac {\sqrt{1 + \frac{1}{x^2}} - 1}{1 + \frac{1}{x}} = \)
- \(= 0 \)
